The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.30 liters. A health campaign promotes the consumption of at least 2.0 liters per day. A sample of 10 adults after the campaign shows the following consumption in liters: 1.361.351.331.661.581.321.381.421.9 01.54
At the 0.050 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.
A.) State the null hypothesis and the alternate hypothesis.
H0: μ ≤
H1: μ >
B.) State the decision rule for 0.050 significance level. (round to 3 decimal places)
C.) Compute the value of the test statistic. (round to 3 decimal places)
D.) At the 0.050 level, can we conclude that water consumption has increased?
E.) Estimate the p-value.

H0: μ ≤ 1.30

H1: μ > 1.30

|Test statistic | > 1.833 ; Reject H0

Test statistic = 3.11

Yes

Pvalue = 0.006

Step-by-step explanation:

H0: μ ≤ 1.30

H1: μ > 1.30

Samples, X ; 1.36,1.35,1.33, 1.66, 1.58, 1.32, 1.38, 1.42, 1.90, 1.54

Xbar = 14.84 / 10 = 1.484

Standard deviation, s = 0.187 (calculator)

Decison rule :

|Test statistic | > TCritical ; reject H0

df = n - 1 = 10 - 1 = 9

Tcritical(0.05; 9) = 1.833

|Test statistic | > 1.833 ; Reject H0

Test statistic :

(xbar - μ) ÷ (s/√(n))

(1.484 - 1.30) ÷ (0.187/√(10))

0.184 / 0.0591345

Test statistic = 3.11

Since ;

|Test statistic | > TCritical ; We reject H0 and conclude that water consumption has increased

Pvalue estimate using the Pvalue calculator :

Pvalue = 0.006

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step-by-step explanation: